BSEC 9-27 - = (0.19)(0.81)^4 = (0.19)(0.4305) = 0.082...

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P(X) = N!/X!(N-X)! * pX(1-P) ^ N-X X= the trial number on which the first defective PC is found 227 20 1. the possibility of asuccess of a single call is p=0.19 2. Let X=the trial number on which the first sale occurs 3. the probability that the “first sale” occurs on trial X is given by P(X)=p(1-p)^x-1 P(5)= (0.19)(1-0.19)^5-1
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Unformatted text preview: = (0.19)(0.81)^4 = (0.19)(0.4305) = 0.082 (8.2%) P(X<=3) =P(1)+P(2)+P(3) P(1)= (.19)(1-.19)^1-1 (.19)(.81)^0 (.19)(1)=.19 P(2)= (.19)(1-.19)^2-1 (.19)(.81)^1 =.1539 P(3)= (.19)(1-.19)^3-1 (.19)(.81)^2 =..1247 P(1)+P(2)+P(3)=.19+.1539+.1247=.4686 (46.86%) P(x>3)=1-P(X<=3) 1-.4686=.5314 (53.14%)...
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This note was uploaded on 12/09/2010 for the course BUS 2313 taught by Professor Zhang during the Fall '10 term at East Central.

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